Question

[Trần Ngọc An Như]

Tìm x biết: \(\frac{x+1}{98}+\frac{x+2}{97}=\frac{x+3}{96}=\frac{x+4}{95}\)

Answer 1

[Trần Việt Linh]

\(\frac{x+1}{98}+\frac{x+2}{97}=\frac{x+3}{96}+\frac{x+4}{95}\)

=> \(\left(\frac{x+1}{98}+1\right)+\left(\frac{x+2}{97}+1\right)=\left(\frac{x+3}{96}+1\right)+\left(\frac{x+4}{95}+1\right)\)

=> \(\frac{x+99}{98}+\frac{x+99}{97}-\frac{x+99}{96}-\frac{x+99}{95}=0\)

=> \(\left(x+99\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

=> \(x+99=0\) (Vì: \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\) )

=>\(x=-99\)

Answer 2

[Trần Nguyễn Bảo Quyên]

Ta có :

\(\frac{x+1}{98}+\frac{x+2}{97}=\frac{x+3}{96}+\frac{x+4}{95}\)

\(\Rightarrow\)  \(\left(\frac{x+1}{98}+1\right)+\left(\frac{x+2}{97}+1\right)=\left(\frac{x+3}{96}+1\right)+\left(\frac{x+4}{95}+1\right)\)

\(\Rightarrow\frac{x+99}{98}+\frac{x+99}{97}=\frac{x+99}{96}+\frac{x+99}{95}\)

\(\Rightarrow\frac{x+99}{98}+\frac{x+99}{97}-\frac{x+99}{96}-\frac{x+99}{95}=0\)

\(\Rightarrow\left(x+99\right).\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

Vì  \(\frac{1}{96}+\frac{1}{97}< \frac{1}{96}+\frac{1}{95}\)

\(\Rightarrow\)  \(\frac{1}{96}+\frac{1}{97}< \frac{1}{96}+\frac{1}{95}\ne0\)

Nên  \(x+99=0\)

\(\Rightarrow x=0-99\)

\(\Rightarrow x=-99\)

Vậy :        \(x=-99\)